Kvant Math Problem 176

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m39s
Source on kvant.digital

Problem

To which side of triangle $ABC$ is the intersection point of its altitudes closest, if $\angle A\lt\angle B\lt\angle C$? And to which vertex?

L. Schneider, 9th-grade student

Exploration

Let $H$ be the orthocenter of triangle $ABC$. The distance from $H$ to side $BC$ equals the length of the perpendicular from $H$ to $BC$. Since $AH$ is perpendicular to $BC$, the foot of this perpendicular is the point where the altitude from $A$ meets $BC$. Hence the distance from $H$ to $BC$ is exactly the segment of the altitude from its foot to $H$.

A standard relation for the orthocenter in an acute triangle is

$$d(H,BC)=CH\cos C=BH\cos B.$$

A more convenient formula is obtained from the circumradius $R$:

$$AH=2R\cos A,\qquad BH=2R\cos B,\qquad CH=2R\cos C.$$

Multiplying by $\sin A$, $\sin B$, $\sin C$ respectively gives

$$d(H,BC)=AH\sin A =2R\sin A\cos A =R\sin 2A,$$

and similarly

$$d(H,CA)=R\sin 2B,\qquad d(H,AB)=R\sin 2C.$$

The problem is then reduced to comparing $\sin 2A$, $\sin 2B$, $\sin 2C$.

A numerical check helps. Take

$$A=40^\circ,\quad B=60^\circ,\quad C=80^\circ.$$

Then

$$\sin 80^\circ\approx0.985,\quad \sin120^\circ\approx0.866,\quad \sin160^\circ\approx0.342.$$

The smallest distance is to side $AB$, opposite the largest angle $C$.

For the nearest vertex, use

$$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C.$$

Since $A<B<C<90^\circ$ in an acute triangle, $\cos A>\cos B>\cos C$, hence

$$AH>BH>CH.$$

Thus the orthocenter is closest to vertex $C$.

The point most likely to hide an error is the comparison of the distances to the sides. One must prove that among $\sin2A,\sin2B,\sin2C$ the smallest is $\sin2C$, using the fact that $A+B+C=180^\circ$ and all angles are acute.

Problem Understanding

We are given a triangle $ABC$ whose angles satisfy

$$\angle A<\angle B<\angle C.$$

We must determine which side is nearest to the orthocenter, the intersection point of the altitudes, and which vertex is nearest to it.

This is a Type C problem, since a nearest side and a nearest vertex must be determined.

The core difficulty is comparing the three distances from the orthocenter to the sides. These distances can be expressed through the angles of the triangle, after which a trigonometric comparison is required.

The expected answer is that the orthocenter is closest to side $AB$, which is opposite the largest angle $C$, and closest to vertex $C$. The reason is that the distances to the sides are proportional to $\sin 2A$, $\sin 2B$, $\sin 2C$, while the distances to the vertices are proportional to $\cos A$, $\cos B$, $\cos C$.

Proof Architecture

The first lemma is that for an acute triangle with circumradius $R$ and orthocenter $H$,

$$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C.$$

This follows from a standard computation in the right triangles formed by the altitudes.

The second lemma is that the distances from $H$ to the sides are

$$d(H,BC)=R\sin2A,\quad d(H,CA)=R\sin2B,\quad d(H,AB)=R\sin2C.$$

This follows by multiplying the altitude lengths from the first lemma by the appropriate sine.

The third lemma is that if

$$A<B<C,\qquad A+B+C=180^\circ,$$

then

$$\sin2C<\sin2B<\sin2A.$$

Since the orthocenter exists inside the triangle only when the triangle is acute, all angles are less than $90^\circ$. Then

$$90^\circ<2C<180^\circ,$$

and $2B,2A$ are smaller than $2C$. The monotonicity of $\sin x$ on $[90^\circ,180^\circ]$ yields the comparison.

The hardest step is proving rigorously that $\sin2C$ is the smallest of the three values.

Solution

Let $H$ be the orthocenter of triangle $ABC$.

Since the orthocenter is the intersection point of the altitudes, the problem concerns an acute triangle. Hence

$$A<B<C<90^\circ.$$

Let $R$ be the circumradius.

Consider the altitude from $A$. If its foot on $BC$ is $D$, then $HD$ is the distance from $H$ to side $BC$. Since $AH\perp BC$, the points $A,H,D$ are collinear and

$$HD=AH-AD.$$

Using the well known relation

$$AH=2R\cos A,$$

we obtain

$$d(H,BC)=AH\sin A =2R\cos A\sin A =R\sin2A.$$

By the same argument,

$$d(H,CA)=R\sin2B,$$

and

$$d(H,AB)=R\sin2C.$$

We now compare these quantities.

Since

$$A<B<C<90^\circ,$$

we have

$$2A<2B<2C<180^\circ.$$

Also,

$$A+B=180^\circ-C>90^\circ,$$

because $C<90^\circ$. Since $B>A$,

$$2B>A+B>90^\circ.$$

Therefore

$$90^\circ<2B<2C<180^\circ.$$

The sine function is strictly decreasing on the interval $[90^\circ,180^\circ]$. Hence

$$\sin2C<\sin2B.$$

Furthermore,

$$2A=360^\circ-2(B+C).$$

Since $B+C>90^\circ$,

$$0^\circ<2A<180^\circ-2B.$$

Using

$$\sin2A = \sin!\bigl(360^\circ-2(B+C)\bigr) = -\sin(2B+2C),$$

and

$$2B+2C>180^\circ,$$

one obtains

$$\sin2A>\sin2B.$$

Consequently,

$$\sin2C<\sin2B<\sin2A.$$

Multiplying by the positive number $R$ gives

$$d(H,AB)<d(H,CA)<d(H,BC).$$

Hence the orthocenter is closest to side $AB$.

Next, compare the distances from $H$ to the vertices. Using

$$AH=2R\cos A,\qquad BH=2R\cos B,\qquad CH=2R\cos C,$$

and the fact that cosine is strictly decreasing on $(0^\circ,90^\circ)$, from

$$A<B<C$$

we obtain

$$AH>BH>CH.$$

Thus the smallest of the three distances is $CH$, so the orthocenter is closest to vertex $C$.

Therefore the nearest side is $AB$ and the nearest vertex is $C$.

$$\boxed{\text{The orthocenter is closest to side }AB\text{ and to vertex }C.}$$

Verification of Key Steps

The first delicate step is the formula

$$d(H,BC)=R\sin2A.$$

Starting from

$$AH=2R\cos A,$$

the distance from $H$ to $BC$ is the component of $AH$ perpendicular to $BC$. Since $AH$ makes angle $A$ with side $AC$,

$$d(H,BC)=AH\sin A.$$

Substituting the expression for $AH$ gives

$$d(H,BC)=2R\cos A\sin A=R\sin2A.$$

A common mistake is to identify the distance with $AH$ itself.

The second delicate step is proving that $\sin2C$ is the smallest value. Since

$$A<B<C<90^\circ,$$

we have

$$2B>90^\circ,\qquad 2C>90^\circ.$$

Both angles lie in the interval where sine decreases, so

$$\sin2C<\sin2B.$$

To compare $\sin2A$ and $\sin2B$, use

$$2A+2B=360^\circ-2C>180^\circ.$$

Since both angles belong to $(0^\circ,180^\circ)$ and $2A<90^\circ<2B$,

$$\sin2A=\sin(180^\circ-2B-2C+180^\circ)>\sin2B.$$

Without checking the location of the angles relative to $90^\circ$, one may incorrectly assume that larger angle implies larger sine.

The third delicate step is the comparison of the distances to the vertices. Because all angles are acute, cosine is positive and strictly decreasing. Hence

$$A<B<C \quad\Longrightarrow\quad \cos A>\cos B>\cos C,$$

which immediately yields

$$AH>BH>CH.$$

If the triangle were not acute, the signs of the cosines would change and the argument would fail.

Alternative Approaches

A synthetic approach uses the feet of the altitudes. Let $D,E,F$ be the feet from $A,B,C$. Then

$$HD=AH\sin A,\qquad HE=BH\sin B,\qquad HF=CH\sin C.$$

Substituting $AH=2R\cos A$ and its analogues again leads to

$$HD:R\sin2A,\quad HE:R\sin2B,\quad HF:R\sin2C.$$

The remainder is the same trigonometric comparison.

Another approach uses barycentric coordinates of the orthocenter,

$$H=(\tan A:\tan B:\tan C).$$

The distances from a point with barycentric coordinates $(x:y:z)$ to the sides are proportional to $x,y,z$. Thus the distances from the orthocenter to the sides are proportional to

$$\tan A,\quad \tan B,\quad \tan C.$$

Converting these proportional distances through the area relations again yields expressions proportional to $\sin2A,\sin2B,\sin2C$. The direct circumradius computation is preferable because it reaches the needed formulas with less machinery.